NCERT Solutions Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

NCERT Solutions for Class 12 Mathematics Chapter 2 is available on the Extramarks website for all kids who are preparing for class 12 board exams. This chapter is about Inverse Trigonometric Functions, and our solutions provide theoretical knowledge and answer solutions to all questions from the NCERT textbook. Our solution is built based on the CBSE NCERT latest 2022-2023 syllabus, and we offer a step-by-step guide for students to comprehend the concepts thoroughly. Students can take full advantage of Extramarks NCERT Solutions for Class 12 Mathematics Chapter 2 for their CBSE Class 12 Term one exam preparation.

The Inverse Trigonometric Functions chapter is an amalgamation of inverse function and trigonometry. It plays a vital role in Calculus concepts to define many integrals. One will learn about the restrictions on domains and ranges of trigonometric functions. Extramarks solutions cover theories, vital concepts, and a lot of solved examples for students to understand the topic easily. If you are looking for perfect study material, one must refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 2.

You can regularly visit the Extramarks website to check the latest updates and news about the CBSE exams. You can also download NCERT Solutions Class 9, NCERT Solutions Class 10, and NCERT Solutions 11, as those would be important for you to build a good foundation for Class 12 preparation.

Key Topics Covered In NCERT Solution for Class 12 Mathematics Chapter 2

To easily understand Class 12 Mathematics Chapter 2, students need to know Chapter 1 on Relations and Functions properly. The underlying principles behind the theories and properties from earlier chapters will continue in Chapter 2. So students should understand functions and how to invert a function and solve questions based on it. The concepts of Inverse trigonometric functions covered in Chapter 2 forms the basis for calculus-based further topics. These concepts of inverse trigonometric functions would go a long way in many science and engineering-oriented curriculum.

In Extramarks NCERT Solutions Class 12 Mathematics Chapter 2, students can expect all topics to be covered and explained in detail. It starts with a chapter introduction and includes sections on basic concepts, trigonometric functions, principle value and inverse function. Studying this chapter will strengthen the foundation of calculus, an important section for many competitive examinations such as NEET and JEE (Main and Advanced) exams.

The key topics covered in chapter 2 include:

Exercise	Topic
2.1	Introduction
2.2	Basic Concepts
2.3	Properties of Inverse Trigonometric Functions
Others	Miscellaneous Q&A

NCERT Solutions Class 12 Mathematics Chapter 2 requires students to use their critical thinking ability and apply a wide range of formulas they have learnt.

2.1 Introduction

This chapter on Inverse trigonometric functions is crucial in higher-grade Mathematics for many engineering and science streams as it builds the fundamentals of calculus. Inverses of trigonometric functions exist to overcome the restrictions we have on domains and their respective ranges. To solve inverse trigonometric functions, one must first understand the trigonometric ratios. The behaviour of the trigonometric functions is usually demonstrated via graphical methods.

2.2 Basic Concepts

This section will quickly revise the corresponding invertible functions. In addition, you will obtain a precise understanding of the functions of graphs and integers. Students will understand the concept that the graph of an inverse function can be extracted from the correlated graph of the original function as a mirror image.

It will develop a better understanding between the domain and the co-domain. The domains are correlated to the principles of functions. This segment will help the students find the unknown angle’s value. Inverse sine and inverse cosine are essential to study. And their basics and root concepts are mentioned in this segment.

2.3 Properties of Trigonometry

This section will cover various essential properties of inverse trigonometric functions. These are the inverse functions of sine, cosine, tangent, cotangent, secant, and cosecant with restricted domains. It would teach students the properties and relationships that exist between various trigonometric functions via formulas.

Few of the formulas covered in the NCERT Solutions for Class 12 Mathematics chapter 2 are given below:

sin-1 (–x) = – sin-1 x
tan-1 x + cot-1 x = π/2
sin-1 x + cos-1 x = π/2
cos-1 (–x) = π – cos-1 x
cot-1 (–x) = π – cot-1 x

Students should also refer to additional study resources such as Extramarks short chapter notes and solved solutions for NCERT Exemplars.

Download NCERT Solutions for Class 12 Mathematics Chapter 2 Exercise & Answer Solutions

NCERT Solutions for Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions is available in a pdf format on the Extramarks website for free. It covers theoretical explanations and has step-by-step solutions to all questions from the NCERT textbook. As the CBSE Class 12 exams approaches, students should leverage the solution guide to get thorough with the Inverse Trigonometric Functions chapter.

In our solutions guide for the Class 12 Mathematics chapter, we have covered topics on domains and ranges of trigonometry-related functions that validate the inverse functions’ existence and help observe their behaviour via graphical representations. Experts advise students to revise this chapter multiple times as it’s a core topic in Calculus that helps define many integrals. And Calculus goes a long way in the advanced level Mathematics syllabus for engineering, architecture and science courses.

Click on the below links to download exercise specific questions and solutions for NCERT Solutions for Class 12 Mathematics Chapter 2:

Chapter 2: Exercise 2.1 Solutions: 14 Questions (12 Short Answers, 2 MCQ)
Chapter 2: Exercise 2.2 Solutions: 21 Questions (18 Short Answers, 3 MCQ)
Chapter 2: Miscellaneous Exercise Solutions: 17 Questions (14 Short Answers, 3 MCQ)

Along with Class 12 Mathematics solutions, you can explore NCERT solutions on our Extramarks website for all primary and secondary classes.

NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions for Class 9, Class 10, aand Class 11.

NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics PDF is an excellent source of information for CBSE students preparing for their 12th standard exams. It consists of various examples with a detailed solution set. While revising from NCERT Exemplars, students gather complete knowledge on the Mathematics concepts and gain deeper insights on a variety of interlinked topics between various chapters.

Exemplar books have proved beneficial for students in CBSE and other curriculum as well. It covers complex theories that confidently prepare students to face the upcoming examinations. Students appearing for Class 12 are advised to download the NCERT Exemplar Class 12 Mathematics pdf file and involve it in their core study materials. Students will get a deeper knowledge of solving complex queries, and it will help prepare them for boards and other entrance examinations.

After referring to the NCERT Solutions and NCERT Exemplar, the students think logically about a problem. As a result, students can jump to more advanced and higher-level conceptual questions. By studying from the Exemplar, you can prepare well for upcoming entrance exams like JEE Main, JEE Advanced, NEET, and WBJEE.

Key Features of NCERT Solutions Class 12 Mathematics Chapter 2

To attain good marks in the exam, students must revise all previous concepts. Therefore, NCERT Solutions for Class 12 Mathematics Chapter 2 offers a complete solution for all problems. The key features are provided:

Extramarks NCERT Solutions Class 12 Mathematics Chapter 2 is prepared by Mathematics experts.
The solutions help to clear the student’s doubts and obstacles by building the base and explaining the chapter’s core concepts.
With the Inverse Trigonometric Functions chapter solutions, students will be able to apply the concepts required to solve many Calculus related problems.

Q.1 Find the principal values of the following

$\sin^{- 1} (- \frac{1}{2})$

Ans.

$\begin{array}{l} Let \sin^{- 1} (- \frac{1}{2}) = x . Then, sinx = - \frac{1}{2} . \\ We know that the range of the principal value branch of \sin^{- 1} is \\ [- \frac{π}{2}, \frac{π}{2}] and \sin (\frac{π}{6}) = \frac{1}{2} . \\ Therefore, principal value of \sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6} \end{array}$

Q.2

$\cos^{- 1} (\frac{\sqrt{3}}{2})$

Ans.

$\begin{array}{l} Let \cos^{- 1} (\frac{\sqrt{3}}{2}) = x . Then, cosx = \frac{\sqrt{3}}{2} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] and \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} . \end{array}$

Q.3

${cosec}^{- 1} (2)$

Ans.

$\begin{array}{l} Let {cosec}^{- 1} (2) = x . Then, cosecx = 2 . \\ We know that the range of the principal value branch of {cosec}^{- 1} is \\ [- \frac{π}{2}, \frac{π}{2}] - {0} and cosec (\frac{π}{6}) = 2 . \\ Therefore, principal value of {cosec}^{- 1} (2) = \frac{π}{6} \end{array}$

Q.4

${tan}^{- 1} (- \sqrt{3})$

Ans.

$\begin{array}{l} Let \tan^{- 1} (- \sqrt{3}) = x . Then, tanx = - \sqrt{3} . \\ We know that the range of the principal value branch of \tan^{- 1} is \\ (- \frac{π}{2}, \frac{π}{2}) and \tan (\frac{π}{3}) = \sqrt{3} . \\ Therefore, principal value of \tan^{- 1} (- \sqrt{3}) = - \frac{π}{3} . \end{array}$

Q.5

${cos}^{- 1} (- \frac{1}{2})$

Ans.

$\begin{array}{l} Let \cos^{- 1} (- \frac{1}{2}) = x . Then, cosx = - \frac{1}{2} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] and \cos (\frac{π}{3}) = \frac{1}{2} . \\ cosx = - \frac{1}{2} \\ = - \cos (\frac{π}{3}) \\ = \cos (π - \frac{π}{3}) \\ = \cos (\frac{2 π}{3}) \\ \Rightarrow x = \frac{2 π}{3} \\ Therefore, principal value of \cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3} . \end{array}$

Q.6

${tan}^{- 1} (- 1)$

Ans.

$\begin{array}{l} Let \tan^{- 1} (- 1) = - 1 . \\ We know that the range of the principle value branch of \tan^{- 1} is \\ (- \frac{π}{2}, \frac{π}{2}) and \tan (\frac{π}{4}) = 1 . \\ Therefore, principle value of \tan^{- 1} (- 1) = - \frac{π}{4} \end{array}$

Q.7

${sec}^{- 1} (\frac{2}{\sqrt{3}})$

Ans.

$\begin{array}{l} Let \sec^{- 1} (\frac{2}{\sqrt{3}}) = x . Then, secx = \frac{2}{\sqrt{3}} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] - \{\frac{π}{2}\} and \sec (\frac{π}{6}) = \frac{2}{\sqrt{3}} . \\ Therefore, principal value of \sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6} \end{array}$

Q.8

${cot}^{- 1} (\sqrt{3})$

Ans.

$\begin{array}{l} Let \cot^{- 1} (\sqrt{3}) = x . Then, cotx = \sqrt{3} . \\ We know that the range of the principal value branch of \cot^{- 1} is \\ (0, π) and \cot (\frac{π}{6}) = \sqrt{3} . \\ Therefore, principal value of \cot^{- 1} (\sqrt{3}) = \frac{π}{6} . \end{array}$

Q.9

${cos}^{- 1} (- \frac{1}{\sqrt{2}})$

Ans.

$\begin{array}{l} Let \cos^{- 1} (- \frac{1}{\sqrt{2}}) = x . Then, cosx = - \frac{1}{\sqrt{2}} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] and \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}} . \\ cosx = - \frac{1}{\sqrt{2}} \\ = - \cos (\frac{π}{4}) \\ = \cos (π - \frac{π}{4}) \\ = \cos (\frac{3 π}{4}) \\ \Rightarrow x = \frac{3 π}{4} \\ Therefore, principal value of \cos^{- 1} (- \frac{1}{2}) = \frac{3 π}{4} . \end{array}$

Q.10

$c o s e c^{- 1} (- \sqrt{2})$

Ans.

$\begin{array}{l} Let {cosec}^{- 1} (- \sqrt{2}) = x . Then, cosecx = - \sqrt{2} . \\ We know that the range of the principal value branch of \\ {cosec}^{- 1} is [- \frac{π}{2}, \frac{π}{2}] - {0} and cosec (\frac{π}{4}) = \sqrt{2} . \\ Therefore, principal value of {cosec}^{- 1} (- \sqrt{2}) = - \frac{π}{4} \end{array}$

Q.11 Find the values of the following:

${tan}^{- 1} (1) + \cos^{- 1} - (\frac{1}{2}) + \sin^{- 1} (- \frac{1}{2})$

Ans.

$\begin{array}{l} Given : \\ \tan^{- 1} (1) + \cos^{- 1} - (\frac{1}{2}) + \sin^{- 1} (- \frac{1}{2}) \\ We shall find the principal value of each term as given below : \\ Let \tan^{- 1} (1) = x . Then, tanx = 1 . \\ We know that the range of the principal value branch of \tan^{- 1} is \\ (- \frac{π}{2}, \frac{π}{2}) and t an (\frac{π}{4}) = 1 . \\ Therefore, principal value of \tan^{- 1} (1) = \frac{π}{4} . \\ second term : \\ Let \cos^{- 1} (- \frac{1}{2}) = x . Then, cosx = - \frac{1}{2} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] and \cos (\frac{π}{3}) = \frac{1}{2} . \\ cosx = - \frac{1}{2} \\ = - \cos (\frac{π}{3}) \\ = \cos (π - \frac{π}{3}) \\ = \cos (\frac{2 π}{3}) \\ and third term : \\ Let \sin^{- 1} (- \frac{1}{2}) = x . Then, sinx = - \frac{1}{2} . \\ We know that the range of the principal value branch of \sin^{- 1} is \\ [- \frac{π}{2}, \frac{π}{2}] and \sin (\frac{π}{6}) = \frac{1}{2} . \\ Therefore, principal value of \sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6} \\ Then, \\ \tan^{- 1} (1) + \cos^{- 1} (- \frac{1}{2}) + \sin^{- 1} (- \frac{1}{2}) = \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} \\ = \frac{3 π + 8 π - 2 π}{12} \\ = \frac{9 π}{12} \\ = \frac{3 π}{4} . \end{array}$

Q.12

${cos}^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2})$

Ans.

$\begin{array}{l} \cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2}) \\ First term : \cos^{- 1} (\frac{1}{2}) \\ Let \cos^{- 1} (\frac{1}{2}) = x . Then, cosx = \frac{1}{2} . \\ We know that the range of the principal value branch of \cos^{- 1} is \\ [0, π] and \cos (\frac{π}{3}) = \frac{1}{2} . \\ Therefore, principal value of \cos^{- 1} (\frac{1}{2}) = \frac{π}{3} \\ Second term : 2 \sin^{- 1} (\frac{1}{2}) \\ Let \sin^{- 1} (\frac{1}{2}) = x . Then, sinx = \frac{1}{2} . \\ We know that the range of the principal value branch of \sin^{- 1} is \\ [- \frac{π}{2}, \frac{π}{2}] and \sin (\frac{π}{6}) = \frac{1}{2} . \\ Therefore, \\ principal value of 2 \sin^{- 1} (\frac{1}{2}) = 2 \times \frac{π}{6} = \frac{π}{3} \\ ∴ \cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2}) = \frac{π}{3} + \frac{π}{3} \\ = \frac{2 π}{3} \end{array}$

Q.13

$\begin{array}{l} {Ifsin}^{- 1} x = y, then \\ A) 0 \leq y \leq π \\ B) - \frac{π}{2} \leq y \leq \frac{π}{2} \\ C) 0 < y < π \\ D) - \frac{π}{2} < y < \frac{π}{2} \end{array}$

Ans.

$\begin{array}{l} Since, \sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}] \\ So, option B is the correct answer . \end{array}$

Q.14

$\begin{array}{l} \tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2) is equal to \\ (A) π \\ (B) - \frac{π}{3} \\ (C) \frac{π}{3} \\ (D) \frac{2 π}{3} \end{array}$

Ans.

$\begin{array}{l} Given : \tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2) \\ Since, \tan^{- 1} \sqrt{3} = \frac{π}{3}, and \\ let \sec^{- 1} (- 2) = x . Then, secx = - 2 . \\ We know that the range of the principal value branch of \sec^{- 1} is \\ [0, π] - \{\frac{π}{2}\} and \sec (\frac{π}{3}) = 2 . \\ ∴ secx = - 2 = - \sec (\frac{π}{3}) \\ = \sec (π - \frac{π}{3}) \\ = \sec (\frac{2 π}{3}) \\ Therefore, principal value of \sec^{- 1} (- 2) = \frac{2 π}{3} . \\ ∴ \tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2) = \frac{π}{3} - (\frac{2 π}{3}) \\ = \frac{π - 2 π}{3} \\ = - \frac{π}{3} . \\ Option B is the correct answer . \end{array}$

Q.15 Prove the following:

${3sin}^{- 1} x = \sin^{- 1} (3 x - 4 x^{3}), x \in [- \frac{1}{2}, \frac{1}{2}]$

Ans.

$\begin{array}{l} R . H . S . = \sin^{- 1} (3 x - 4 x^{3}) \\ = \sin^{- 1} (3 sinθ - 4 \sin^{3} θ) [Putting x = sinθ] \\ = \sin^{- 1} (\sin 3 θ) [∵ \sin 3 x = 3 sinx - 4 \sin^{3} x] \\ = 3 θ \\ = 3 \sin^{- 1} x [∴ x = sinθ \Rightarrow θ = \sin^{- 1} x] \\ = R . H . S . \end{array}$

Q.16

${3cos}^{- 1} x = c o s^{- 1} (4 x^{3} - 3 x), x \in [\frac{1}{2}, 1]$

Ans.

$\begin{array}{l} R . H . S . = \cos^{- 1} (4 x^{3} - 3 x) \\ = \cos^{- 1} (4 \cos^{3} θ - 3 cosθ) [Putting x = cosθ] \\ = \cos^{- 1} (\cos 3 θ) [∵ \cos 3 x = 4 \cos^{3} x - 3 cosx] \\ = 3 θ \\ = 3 \cos^{- 1} x [∴ x = cosθ \Rightarrow θ = \cos^{- 1} x] \\ = R . H . S . \end{array}$

Q.17

${tan}^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) = \tan^{- 1} (\frac{1}{2})$

Ans.

$\begin{array}{l} L . H . S . = \tan^{- 1} (\frac{2}{11}) + \tan^{- 1} (\frac{7}{24}) \\ = \tan^{- 1} (\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} (\frac{\frac{48 + 77}{11 \times 24}}{\frac{11 \times 24 - 14}{11 \times 24}}) \\ = \tan^{- 1} (\frac{125}{250}) \\ = \tan^{- 1} (\frac{1}{2}) = R . H . S . \end{array}$

Q.18

${2tan}^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})$

Ans.

$\begin{array}{l} L . H . S = 2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) \\ = \tan^{- 1} (\frac{2 \times \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}}) + \tan^{- 1} (\frac{1}{7}) [∵ 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})] \\ = \tan^{- 1} (\frac{4}{3}) + \tan^{- 1} (\frac{1}{7}) \\ = \tan^{- 1} (\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}) [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} (\frac{31}{17}) \\ = R . H . S . \end{array}$

Q.19 Write the following functions in the simplest form:

${tan}^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}, x \neq 0$

Ans.

$\begin{array}{l} Given : \\ \tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x} = \tan^{- 1} \frac{\sqrt{1 + \tan^{2} θ} - 1}{tanθ} [Putting x = tanθ] \\ = \tan^{- 1} \frac{\sqrt{\sec^{2} θ} - 1}{tanθ} [∵ 1 + \tan^{2} θ = \sec^{2} θ] \\ = \tan^{- 1} \frac{secθ - 1}{tanθ} \\ = \tan^{- 1} \frac{1 - cosθ}{sinθ} \\ = \tan^{- 1} (\tan (\frac{θ}{2})) [∵ \tan^{- 1} (tanx) = x] \\ = \frac{θ}{2} \\ = \frac{1}{2} \tan^{- 1} x \end{array}$

Q.20

${tan}^{- 1} \frac{1}{\sqrt{x^{2} - 1}}, | x | > 1$

Ans.

${tan}^{- 1} \frac{1}{\sqrt{x^{2} - 1}}, | x | > 1$

Q.21

t a n^{- 1} (\sqrt{\frac{1 - c o s x}{1 + c o s x}}), x < π

Ans.

$\begin{array}{l} {tan}^{-1} (\sqrt{\frac{1-cosx}{1+cosx}}) = {tan}^{-1} (\sqrt{\frac{{2sin}^{2} (\frac{x}{2})}{{2cos}^{2} (\frac{x}{2})}}) [\begin{array}{l} ∵ \cos 2 x = 1 - 2 \sin^{2} x \\ = 2 \cos^{2} x - 1 \end{array}] \\ = \tan^{- 1} (\tan \frac{x}{2}) \\ = \frac{x}{2} [∵ \tan^{- 1} (\tan x) = x] \end{array}$

Q.22

${tan}^{- 1} (\frac{cosx - sinx}{cosx + sinx}), 0 < x < π$

Ans.

$\begin{array}{l} \tan^{- 1} (\frac{cosx - sinx}{cosx + sinx}) = \tan^{- 1} (\frac{1 - tanx}{1 + tanx}) [\begin{array}{l} Dividing numerator \\ and denominator by sinx \end{array}] \\ = \tan^{- 1} \{\tan (\frac{π}{4} - x)\} [∵ \tan (\frac{π}{4} - x) = \frac{1 - tanx}{1 + tanx}] \\ = \frac{π}{4} - x [∵ \tan^{- 1} (tanx) = x] \end{array}$

Q.23

${tan}^{- 1} \frac{x}{\sqrt{a^{2} - x^{2}}}, | x | < a$

Ans.

$\begin{array}{l} \tan^{- 1} \frac{x}{\sqrt{a^{2} - x^{2}}} = \tan^{- 1} \frac{asinθ}{\sqrt{a^{2} - a^{2} \sin^{2} θ}} [Putting x = a \sin θ] \\ = \tan^{- 1} (\frac{asinθ}{acosθ}) [∵ 1 - \sin^{2} θ = \cos^{2} θ] \\ = \tan^{- 1} (tanθ) [\frac{sinθ}{cosθ} = tanθ] \\ = θ \\ = \sin^{- 1} (\frac{x}{a}) [∵ x = asinθ] \end{array}$

Q.24

${tan}^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}), a > 0; - \frac{a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$

Ans.

$\begin{array}{l} \tan^{- 1} (\frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}) = \tan^{- 1} (\frac{3 a^{3} tanθ - {(atanθ)}^{3}}{a^{3} - 3 a {(atanθ)}^{2}}) [Putting x = atanθ] \\ = \tan^{- 1} (\frac{3 a^{3} tanθ - a^{3} \tan^{3} θ}{a^{3} - 3 a^{3} \tan^{2} θ}) \\ = \tan^{- 1} \{\frac{a^{3} (3 tanθ - \tan^{3} θ)}{a^{3} (1 - 3 \tan^{2} θ)}\} \\ = \tan^{- 1} (\tan 3 θ) [\tan 3 θ = \frac{3 tanθ - \tan^{3} θ}{1 - 3 \tan^{2} θ}] \\ = 3 θ \\ = 3 \tan^{- 1} (\frac{x}{a}) [∵ x = atanθ] \end{array}$

Q.25 Find the values of each of the following:

${tan}^{- 1} \{2 \cos (2 \sin^{- 1} \frac{1}{2})\}$

Ans.

$\begin{array}{l} \tan^{- 1} \{2 \cos (2 \sin^{- 1} \frac{1}{2})\} \tan^{- 1} \{2 \cos (2 \times \frac{π}{6})\} \\ = \tan^{- 1} \{2 \cos (\frac{π}{3})\} \\ = \tan^{- 1} \{2 \times \frac{1}{2}\} \\ = \tan^{- 1} (1) = \frac{π}{4} \end{array}$

Q.26

$cot (\tan^{- 1} a + \cot^{- 1} a)$

Ans.

$\begin{array}{l} \cot (\tan^{- 1} a + \cot^{- 1} a) = \cot [∵ \tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}] (\frac{π}{2}) \\ = 0 \end{array}$

Q.27

$tan \frac{1}{2} (\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}), | x | < 1, y > 0 and xy < 1$

Ans.

$\begin{array}{l} \tan \frac{1}{2} (\sin^{- 1} \frac{2 x}{1 + x^{2}} + \cos^{- 1} \frac{1 - y^{2}}{1 + y^{2}}) \\ = \tan \frac{1}{2} (\sin^{- 1} \frac{2 tanα}{1 + \tan^{2} α} + \cos^{- 1} \frac{1 - \tan^{2} β}{1 + \tan^{2} β}) [Putting x = tanα, y = tanβ] \\ = \tan \frac{1}{2} \{\sin^{- 1} (\sin 2 α) + \cos^{- 1} (\cos 2 β)\} \\ = \tan \frac{1}{2} (2 α + 2 β) \\ = \tan (α + β) \\ = \frac{tanα + tanβ}{1 - tanαtanβ} \\ = \frac{x + y}{1 - xy} \end{array}$

Q.28

$If \sin (\sin^{- 1} \frac{1}{5} + \cos^{- 1} x) = 1, then find the value of x .$

Ans.

$\begin{array}{l} \sin (\sin^{- 1} \frac{1}{5} + \cos^{- 1} x) = 1 \\ \Rightarrow \sin (\sin^{- 1} \frac{1}{5} + \cos^{- 1} x) = \sin^{- 1} (\frac{π}{2}) \\ \Rightarrow \sin^{- 1} \frac{1}{5} + \cos^{- 1} x = \frac{π}{2} \\ \Rightarrow \sin^{- 1} \frac{1}{5} = \frac{π}{2} - \cos^{- 1} x \\ \Rightarrow \sin^{- 1} \frac{1}{5} = \sin^{- 1} x [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}] \\ \Rightarrow \frac{1}{5} = x \end{array}$

Q.29

${If tan}^{- 1} \frac{x - 1}{x - 2} + \tan^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}, then find the value of x .$

Ans.

$\begin{array}{l} \tan^{- 1} \frac{x - 1}{x - 2} + \tan^{- 1} \frac{x +1}{x +2} = \frac{π}{4} \\ \Rightarrow \tan^{- 1} (\frac{\frac{x - 1}{x - 2} + \frac{x +1}{x +2}}{1 - (\frac{x - 1}{x - 2}) (\frac{x +1}{x +2})}) = \frac{π}{4} [∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ \Rightarrow (\frac{\frac{x^{2} + x - 2 + x^{2} - x - 2}{(x - 2) (x +2)}}{\frac{x^{2} - 4 - x^{2} + 1}{(x - 2) (x +2)}}) = \tan \frac{π}{4} \\ \Rightarrow (\frac{2 x^{2} - 4}{- 3}) = 1 \\ \Rightarrow 2 x^{2} - 4 = - 3 \\ \Rightarrow x = \pm \frac{1}{\sqrt{2}} \end{array}$

Q.30

${sin}^{- 1} (\sin \frac{2 π}{3})$

Ans.

$\begin{array}{l} \sin^{- 1} (\sin \frac{2 π}{3}) = \sin ∵ Range of sinθ is {[- \frac{π}{2}, \frac{π}{2}]}^{- 1} \{\sin (π - \frac{π}{3})\} \\ = \sin^{- 1} \{\sin \frac{π}{3}\} [∵ \sin (π - θ) = sinθ] \\ = \frac{π}{3} \end{array}$

Q.31

${tan}^{- 1} (t a n \frac{3 π}{4})$

Ans.

$\begin{array}{l} \tan^{- 1} (\tan \frac{3 π}{4}) = \tan^{- 1} \{\tan (π - \frac{π}{4})\} [∵ Range of tanθ is (- \frac{π}{2}, \frac{π}{2})] \\ = \tan^{- 1} \{- \tan \frac{π}{4}\} [∵ \tan (π - θ) = - tanθ] \\ = \tan^{- 1} \{\tan (- \frac{π}{4})\} [\tan (- θ) = - tanθ] \\ = - \frac{π}{4} \end{array}$

Q.32

$tan (\sin^{- 1} \frac{3}{5} + \cot^{- 1} \frac{3}{2})$

Ans.

$\begin{array}{l} \tan (\sin^{- 1} \frac{3}{5} + \cot^{- 1} \frac{3}{2}) \\ Let \sin^{- 1} \frac{3}{5} = x and \cot^{- 1} \frac{3}{2} = y \\ \Rightarrow sinx = \frac{3}{5} and coty = \frac{3}{2} \\ \Rightarrow cosx = \sqrt{1 - \sin^{2} x} \\ = \sqrt{1 - {(\frac{3}{5})}^{2}} = \frac{4}{5} \\ and \tan y = \frac{1}{\cot y} \\ = \frac{1}{(\frac{3}{2})} = \frac{2}{3} \\ \Rightarrow tanx = \frac{sinx}{cosx} = \frac{(\frac{3}{5})}{(\frac{4}{5})} \\ = \frac{3}{4} \\ \Rightarrow x = \tan^{- 1} (\frac{3}{4}) and y = \tan^{- 1} (\frac{2}{3}) \\ \Rightarrow \sin^{- 1} \frac{3}{5} = \tan^{- 1} (\frac{3}{4}) and \cot^{- 1} \frac{3}{2} = \tan^{- 1} (\frac{2}{3}) \\ ∴ \tan (\sin^{- 1} \frac{3}{5} + \cot^{- 1} \frac{3}{2}) = \tan \{\tan^{- 1} (\frac{3}{4}) + \tan^{- 1} (\frac{2}{3})\} \\ = \tan \{\tan^{- 1} (\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}})\} \\ = \tan \{\tan^{- 1} (\frac{17}{6})\} \\ = \frac{17}{6} \end{array}$

Q.33

$\begin{array}{l} \cos^{- 1} (\cos \frac{7 π}{6}) is equal to \\ (A) \frac{7 π}{6} \\ (B) \frac{5 π}{6} \\ (C) \frac{π}{3} \\ (D) \frac{π}{6} \end{array}$

Ans.

$\begin{array}{l} \cos^{- 1} (\cos \frac{7 π}{6}) = \cos^{- 1} \{\cos (2 π - \frac{5 π}{6})\} [∵ Range of cosθ is [0, π]] \\ = \cos^{- 1} \{\cos (\frac{5 π}{6})\} [∵ \cos (2 π - θ) = cosθ] \\ = \frac{5 π}{6} [∵ \cos^{- 1} (cosx) = x] \\ Hence option (b) is correct . \end{array}$

Q.34

$\begin{array}{l} \sin (\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})) is equal to \\ (A) \frac{1}{2} \\ (B) \frac{1}{3} \\ (C) \frac{1}{4} \\ (D) 1 \end{array}$

Ans.

$\begin{array}{l} \sin (\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})) = \sin (\frac{π}{3} + \sin^{- 1} (\frac{1}{2})) [\sin (- θ) = - sinθ] \\ = \sin (\frac{π}{3} + \frac{π}{6}) [∵ \sin \frac{π}{6} = \frac{1}{2}] \\ = \sin (\frac{2 π + π}{6}) \\ = \sin (\frac{3 π}{6}) \\ = \sin (\frac{π}{2}) \\ = 1 [∵ \sin \frac{π}{2} = 1] \\ Hence, option (D) is correct . \end{array}$

Q.35

$\begin{array}{l} \tan^{- 1} \sqrt{3} - \cot^{- 1} (- \sqrt{3}) is equal to \\ \begin{array}{l} (A) π \\ (B) - \frac{π}{2} \\ (C) 0 \\ (D) 2 \sqrt{3} \end{array} \end{array}$

Ans.

$\begin{array}{l} \tan^{- 1} \sqrt{3} - \cot^{- 1} (- \sqrt{3}) \\ = - \tan^{- 1} (- \sqrt{3}) - \cot^{- 1} (- \sqrt{3}) [∵ \tan (- θ) = tanθ] \\ = - {\tan^{- 1} (- \sqrt{3}) + \cot^{- 1} (- \sqrt{3})} \\ = - \frac{π}{2} [∵ \tan^{- 1} x + {con}^{- 1} x = \frac{π}{2}] \\ Hence option (B) is correct . \end{array}$

Q.36

${Find the value of cos}^{- 1} (\cos \frac{13 π}{6}) .$

Ans.

$\begin{array}{l} Range of cosθ is [0, π] \\ ∴ \cos^{- 1} (\cos \frac{13 π}{6}) = \cos^{- 1} \{\cos (2 π + \frac{π}{6})\} \\ = \cos^{- 1} \{\cos (\frac{π}{6})\} [∵ \cos (2 π + θ) = \cos θ] \\ = \frac{π}{6} \end{array}$

Q.37

${Find the value of tan}^{- 1} (\tan \frac{7 π}{6}) .$

Ans.

$\begin{array}{l} Range of tanθ is (- \frac{π}{2}, \frac{π}{2}) \\ ∴ \tan^{- 1} (\tan \frac{7 π}{6}) = \tan^{- 1} \{\tan (π + \frac{π}{6})\} \\ = \tan^{- 1} \{(\tan \frac{π}{6})\} [∵ \tan (π + θ) = tanθ] \\ = \frac{π}{6} \end{array}$

Q.38

$\begin{array}{l} Prove that : \\ 2 \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} \frac{24}{7} \end{array}$

Ans.

$\begin{array}{l} Let \sin^{- 1} (\frac{3}{5}) = x \Rightarrow sinx = \frac{3}{5} \\ and cosx = \sqrt{1 - \sin^{2} x} \\ = \sqrt{1 - {(\frac{3}{5})}^{2}} \\ = \frac{4}{5} \\ tanx = \frac{sinx}{cosx} \\ = \frac{(\frac{3}{5})}{(\frac{4}{5})} \\ = \frac{3}{4} \\ Since, \\ \tan 2 x = \frac{2 \tan x}{1 + \tan^{2} x} \\ = \frac{2 \times \frac{3}{4}}{1 - {(\frac{3}{4})}^{2}} \\ = \frac{24}{17} \\ 2 x = \tan^{- 1} (\frac{24}{17}) \\ 2 \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{24}{17}) \end{array}$

Q.39

$\begin{array}{l} Prove that : \\ \sin^{- 1} (\frac{8}{17}) + \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} \frac{77}{36} \end{array}$

Ans.

$\begin{array}{l} \sin^{- 1} (\frac{8}{17}) + \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} \frac{77}{36} \\ Let \sin^{- 1} (\frac{8}{17}) = x \Rightarrow sinx = \frac{8}{17} \\ and cosx = \sqrt{1 - \sin^{2} x} \\ = \sqrt{1 - {(\frac{8}{17})}^{2}} \\ = \frac{15}{17} \\ tanx = \frac{sinx}{cosx} \\ = \frac{(\frac{8}{17})}{(\frac{15}{17})} \\ = \frac{8}{15} \\ \Rightarrow x = \tan^{- 1} \frac{8}{15} \\ \Rightarrow \sin^{- 1} (\frac{8}{17}) = \tan^{- 1} \frac{8}{15} \\ Again, let \sin^{- 1} (\frac{3}{5}) = y \Rightarrow siny = \frac{3}{5} \\ so, tany = \frac{siny}{\sqrt{1 - \sin^{2} y}} \\ = \frac{(\frac{3}{5})}{\sqrt{1 - {(\frac{3}{5})}^{2}}} \\ = \frac{3}{4} \\ y = \tan^{- 1} (\frac{3}{4}) \\ ∴ \sin^{- 1} (\frac{3}{5}) = \tan^{- 1} (\frac{3}{4}) \\ L . H . S = \sin^{- 1} (\frac{8}{17}) + si n^{- 1} (\frac{3}{5}) \\ = \tan^{- 1} (\frac{8}{15}) + \tan^{- 1} (\frac{3}{4}) \\ = \tan^{- 1} (\frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}}) \\ = \tan^{- 1} (\frac{77}{36}) = R . H . S . \end{array}$

Q.40

$\begin{array}{l} Prove that : \\ \cos^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{12}{13}) = \cos^{- 1} \frac{33}{65} \end{array}$

Ans.

$\begin{array}{l} Let \cos^{- 1} (\frac{4}{5}) = x and \cos^{- 1} (\frac{12}{13}) = y \\ \Rightarrow cosx = \frac{4}{5} and cosy = \frac{12}{13} \\ \Rightarrow sinx = \sqrt{1 - \cos^{2} x} and siny = \sqrt{1 - \cos^{2} y} \\ \Rightarrow sinx = \sqrt{1 - {(\frac{4}{5})}^{2}} and siny = \sqrt{1 - {(\frac{12}{13})}^{2}} \\ \Rightarrow sinx = \frac{3}{5} and siny = \frac{5}{13} \\ Then, \\ \cos (x + y) = cosx cosy - sinx siny \\ = \frac{4}{5} \times \frac{12}{13} - \frac{3}{5} \times \frac{5}{13} \\ = \frac{48 - 15}{5 \times 13} \\ \Rightarrow x + y = \cos^{- 1} (\frac{33}{65}) \\ \Rightarrow \cos^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{12}{13}) = \cos^{- 1} \frac{33}{65} . \end{array}$

Q.41

$\begin{array}{l} Prove that : \\ \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} \frac{56}{65} \end{array}$

Ans.

$\begin{array}{l} Let \cos^{- 1} (\frac{12}{13}) = x and \sin^{- 1} (\frac{3}{5}) = y \\ \Rightarrow cosx = \frac{12}{13} and siny = \frac{3}{5} \\ \Rightarrow sinx = \sqrt{1 - \cos^{2} x} and cosy = \sqrt{1 - \sin^{2} y} \\ \Rightarrow sinx = \sqrt{1 - {(\frac{12}{13})}^{2}} and cosy = \sqrt{1 - {(\frac{3}{5})}^{2}} \\ \Rightarrow sinx = \frac{5}{13} and cosy = \frac{4}{5} \\ Since, \\ \sin (x + y) = sinx cosy + siny cosx \\ = (\frac{5}{13}) (\frac{4}{5}) + (\frac{3}{5}) (\frac{12}{13}) \\ = \frac{20 + 36}{65} \\ = \frac{56}{65} \\ \Rightarrow x + y = \sin^{- 1} (\frac{56}{65}) \\ \Rightarrow \cos^{- 1} (\frac{12}{13}) + \sin^{- 1} (\frac{3}{5}) = \sin^{- 1} (\frac{56}{65}) . \end{array}$

Q.42

$\begin{array}{l} Prove that : \\ \tan^{- 1} (\frac{63}{16}) = \sin^{- 1} (\frac{5}{13}) + \cos^{- 1} \frac{3}{5} \end{array}$

Ans.

$\begin{array}{l} Let \sin^{- 1} (\frac{5}{13}) = x and \cos^{- 1} \frac{3}{5} = y \\ \Rightarrow sinx = \frac{5}{13} and cosy = \frac{3}{5} \\ \Rightarrow cosx = \sqrt{1 - \sin^{2} x} and siny = \sqrt{1 - \cos^{2} y} \\ \Rightarrow cosx = \sqrt{1 - {(\frac{5}{13})}^{2}} and siny = \sqrt{1 - {(\frac{3}{5})}^{2}} \\ \Rightarrow cosx = \frac{12}{13} and siny = \frac{4}{5} \\ \Rightarrow tanx = \frac{sinx}{cosx} = \frac{5}{12} and tany = \frac{4}{3} \\ Since \tan (x + y) = \frac{tanx + tany}{1 - tanxtany} \\ = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}} \\ \Rightarrow x + y = \tan^{- 1} (\frac{63}{16}) \\ \Rightarrow \sin^{- 1} (\frac{5}{13}) + \cos^{- 1} \frac{3}{5} = \tan^{- 1} (\frac{63}{16}) \end{array}$

Q.43

$\begin{array}{l} Prove that : \\ \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{8}) = \frac{π}{4} \end{array}$

Ans.

$\begin{array}{l} L . H . S . = \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{8}) \\ = \tan^{- 1} (\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}}) + \tan^{- 1} (\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}}) \\ [\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})] \\ = \tan^{- 1} (\frac{12}{34}) + \tan^{- 1} (\frac{11}{23}) \\ = \tan^{- 1} (\frac{\frac{12}{34} + \frac{11}{23}}{1 - \frac{12}{34} \times \frac{11}{23}}) \\ = \tan^{- 1} (\frac{650}{650}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} = R . H . S . \end{array}$

Q.44

$\begin{array}{l} Prove that : \\ \tan^{- 1} \sqrt{x} = \frac{1}{2} \cos^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1] \end{array}$

Ans.

$\begin{array}{l} R . H . S . = \frac{1}{2} \cos^{- 1} (\frac{1 - x}{1 + x}) \\ = \frac{1}{2} \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) [Putting x = \tan^{2} θ] \\ = \frac{1}{2} \cos^{- 1} (\cos 2 θ) \\ = 2 θ \\ = 2 \tan^{- 1} \sqrt{x} [since, tanθ = \sqrt{x} \Rightarrow θ = \tan^{- 1} \sqrt{x}] \\ = L . H . S . \end{array}$

Q.45

$\begin{array}{l} Prove that : \\ \cot^{- 1} (\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}) = \frac{x}{2}, x \in (0, \frac{π}{4}) \end{array}$

Ans.

$\begin{array}{l} L . H . S . \\ = \cot^{- 1} (\frac{\sqrt{1 + sinx} + \sqrt{1 - sinx}}{\sqrt{1 + sinx} - \sqrt{1 - sinx}}) \\ = \cot^{- 1} (\frac{\sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}} + \sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}}{\sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}} - \sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}}) \\ = \cot^{- 1} (\frac{\sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}} + \sqrt{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}}{\sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}} - \sqrt{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}}) \\ = \cot^{- 1} (\frac{\sin \frac{x}{2} + \cos \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2} - (\cos \frac{x}{2} - \sin \frac{x}{2})}) \\ = \cot^{- 1} (\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}) \\ = \cot^{- 1} (\cot \frac{x}{2}) = \frac{x}{2} = R . H . S . \end{array}$

Q.46

$\begin{array}{l} Prove that : \\ \tan^{- 1} (\frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x, - \frac{1}{\sqrt{2}} \leq x \leq 1 \\ [Hint : Putx = \cos 2 θ] \end{array}$

Ans.

$\begin{array}{l} L . H . S . = \tan^{- 1} (\frac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}}) \\ = \tan^{- 1} (\frac{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ} - \sqrt{1 - \cos 2 θ}}) [Putting x = \cos 2 θ] \\ = \tan^{- 1} (\frac{\sqrt{2 \cos^{2} θ} + \sqrt{2 \sin^{2} θ}}{\sqrt{2 \cos^{2} θ} - \sqrt{2 \sin^{2} θ}}) \\ = \tan^{- 1} (\frac{cosθ + sinθ}{cosθ - sinθ}) \\ = \tan^{- 1} (\frac{1 + \frac{sinθ}{cosθ}}{1 - \frac{sinθ}{cosθ}}) \\ = \tan^{- 1} (\frac{1 + tanθ}{1 - tanθ}) \\ = \tan^{- 1} \tan (\frac{π}{4} - θ) = \frac{π}{4} - θ = R . H . S . \end{array}$

Q.47

$\begin{array}{l} Prove that : \\ \frac{9 π}{8} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) = \frac{9}{4} \sin^{- 1} (\frac{2 \sqrt{2}}{3}) \end{array}$

Ans.

$\begin{array}{l} L . H . S . = \frac{9 π}{8} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) \\ = \frac{9}{4} {\frac{π}{2} - \sin^{- 1} (\frac{1}{3})} \\ = \frac{9}{4} \cos^{- 1} (\frac{1}{3}) . .. (i) \\ Now, let x = \cos^{- 1} (\frac{1}{3}) \\ \Rightarrow cosx = \frac{1}{3} \\ \Rightarrow sinx = \sqrt{1 - \cos^{2} x} \\ = \sqrt{1 - {(\frac{1}{3})}^{2}} \\ = \frac{2 \sqrt{2}}{3} \\ x = \sin^{- 1} (\frac{2 \sqrt{2}}{3}) \\ ∴ \cos^{- 1} (\frac{1}{3}) = \sin^{- 1} (\frac{2 \sqrt{2}}{3}) \\ So, from eq (i), we get \\ L . H . S . = \frac{9}{4} \cos^{- 1} (\frac{1}{3}) \\ = \frac{9}{4} \sin^{- 1} (\frac{2 \sqrt{2}}{3}) = R . H . S . \end{array}$

Q.48

Ans.

Q.49

$\begin{array}{l} Prove that : \\ \frac{9 π}{8} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) = \frac{9}{4} \sin^{- 1} (\frac{2 \sqrt{2}}{3}) \end{array}$

Ans.

Q.50 Solve 2tan^-1(cosx) = tan^-1(2cosecx)

Ans.

$\begin{array}{l} {2tan}^{-1} (cosx) = {tan}^{-1} (2cosecx) \\ \Rightarrow {tan}^{-1} (\frac{2cosx}{1 - \cos^{2} x}) = {tan}^{-1} (\frac{2}{\sin x}) [∵ 2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})] \\ \Rightarrow \frac{2cosx}{1 - \cos^{2} x} = \frac{2}{\sin x} \\ \Rightarrow \frac{cosx}{\sin^{2} x} = \frac{1}{\sin x} \\ \Rightarrow \cos x = \sin x \\ \Rightarrow 1 = \tan x \\ \Rightarrow \tan \frac{π}{4} = \tan x \\ ∴ x = \frac{π}{4} \end{array}$

Q.51

$\begin{array}{l} S o l v e : \\ t a n^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} t a n^{- 1} x, (x > 0) \end{array}$

Ans.

$\begin{array}{l} \tan^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{- 1} x \\ \Rightarrow 2 \tan^{- 1} \frac{1 - x}{1 + x} = \tan^{- 1} x \\ \Rightarrow \tan^{- 1} {\frac{2 (\frac{1 - x}{1 + x})}{1 - {(\frac{1 - x}{1 + x})}^{2}}} = \tan^{- 1} x \\ \Rightarrow \frac{2 (\frac{1 - x}{1 + x})}{1 - {(\frac{1 - x}{1 + x})}^{2}} = x \\ \Rightarrow \frac{2 (1 - x) (1 + x)}{{(1 + x)}^{2} - {(1 - x)}^{2}} = x \\ \Rightarrow \frac{2 (1 - x^{2})}{4 x} = x \\ \Rightarrow 1 - x^{2} = 2 x^{2} \\ \Rightarrow 1 = 3 x^{2} \\ \Rightarrow x = \pm \frac{1}{\sqrt{3}} \end{array}$

Q.52

$\begin{array}{l} \sin (\tan^{- 1} x), |x| < 1 is equal to \\ (a) \frac{x}{\sqrt{1 - x^{2}}} \\ (b) \frac{1}{\sqrt{1 - x^{2}}} \\ (c) \frac{1}{\sqrt{1 + x^{2}}} \\ (d) \frac{x}{\sqrt{1 + x^{2}}} \end{array}$

Ans.

$\begin{array}{l} sin ({tan}^{-1} x) \\ L e t {tan}^{-1} x = θ \Rightarrow x = \tan θ \\ \Rightarrow \sec^{2} θ = 1 + \tan^{2} θ \\ \Rightarrow \frac{1}{\cos^{2} θ} = 1 + x^{2} \\ \Rightarrow \cos^{2} θ = \frac{1}{1 + x^{2}} \\ S i n c e, \\ \Rightarrow \sin θ = \sqrt{1 - \cos^{2} θ} \\ = \sqrt{1 - \frac{1}{1 + x^{2}}} \\ = \sqrt{\frac{1 + x^{2} - 1}{1 + x^{2}}} = \frac{x}{\sqrt{1 + x^{2}}} \\ θ = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}) \\ \Rightarrow {tan}^{-1} x = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}) \\ T h e r e f o r e, \\ sin ({tan}^{-1} x) = \sin {\sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}})} \\ = \frac{x}{\sqrt{1 + x^{2}}} \\ T h u s, option ‘D’ is correct solution . \end{array}$

Q.53

$\begin{array}{l} \sin^{- 1} (1 - x) - 2 \sin^{- 1} x = \frac{π}{2}, then x is equal to \\ (A) 0, \frac{1}{2} \\ (B 1, \frac{1}{2} \\ (C) 0 \\ (D) \frac{1}{2} \end{array}$

Ans.

$\begin{array}{l} {sin}^{-1} (1 - x) - {2sin}^{-1} x = \frac{π}{2} \\ 1 - x = \sin (\frac{π}{2} + 2 \sin^{- 1} x) \\ = \cos (2 \sin^{- 1} x) [∵ s i n (\frac{π}{2} + x) = \cos x] \\ = \cos {\cos^{- 1} (1 - 2 x^{2})} (\begin{array}{l} L e t \sin^{- 1} x = θ and cos2 θ =1 - {2sin}^{2} x \\ 2 θ = \cos^{- 1} (1 - 2 \sin^{2} x) \\ \Rightarrow 2 \sin^{- 1} x = \cos^{- 1} (1 - x^{2}) \end{array}) \\ \Rightarrow 1 - x = 1 - 2 x^{2} \\ \Rightarrow 2 x^{2} - x = 0 \\ \Rightarrow x (2 x - 1) = 0 \\ \Rightarrow x = 0 or 2x - 1 = 0 \\ \Rightarrow x = 0 or x = \frac{1}{2} \\ But on substitution of x = \frac{1}{2} \\ L . H . S . = {sin}^{-1} (1-x) - {2sin}^{-1} x \\ = {sin}^{- 1} (1 - \frac{1}{2}) - {2sin}^{-1} (\frac{1}{2}) \\ = \frac{π}{6} - 2 (\frac{π}{6}) \\ = - \frac{π}{6} \neq \frac{π}{2} \neq R . H . S . \\ ∴ x = \frac{1}{2} is not the solution of given equation . \\ Thus, x = 0 is the solution of given equation . \\ Therefore, option ‘C’ is correction solution of given equation . \end{array}$

Q.54

$\begin{array}{l} \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y}) is equal to \\ (A) \frac{π}{2} \\ (B) \frac{π}{3} \\ (C) \frac{π}{4} \\ (D) \frac{- 3 π}{4} \end{array}$

Ans.

$\begin{array}{l} \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y}) \\ = \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{\frac{x}{y} - 1}{\frac{x}{y} + 1}) \\ = \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} {\frac{- (1 - \frac{x}{y})}{(1 + \frac{x}{y})}} \\ = \tan^{- 1} (\frac{x}{y}) + \tan^{- 1} {\frac{(1 - \frac{x}{y})}{(1 + \frac{x}{y})}} [∵ \tan^{- 1} (- x) = - \tan^{- 1} x] \\ = \tan^{- 1} (\frac{x}{y}) + \tan^{- 1} (1) - \tan^{- 1} (\frac{x}{y}) \\ = \tan^{- 1} (1) \\ = \frac{π}{4} \\ Therefore, option ‘ C ‘ is correct solution . \end{array}$

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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One can get complete NCERT Solutions Class 12 Mathematics Chapter 2 on the Extramark website. Along with class 12, one can find NCERT solutions right from class 1 to class 12 on our website.

2. How many questions are included in NCERT Solutions for Class 12 Mathematics Chapter 2?

A total of 52 questions are there in Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions. These are spread across three exercises: basics, principles of trigonometry, and a miscellaneous one. The problems cover various topics and formulas from the chapter and vary in difficulty level. Mathematics professors advise students to devote a reasonable amount of time to solve these questions as this chapter forms a base for future Calculus topics.

3. What are inverse trigonometric functions?

The inverse trigonometric functions help to determine angles and trigonometric ratios. The functions include sin-1 x, cos-1 x, tan-1 x, cot-1 x, cosec-1 x, sec-1 x.

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

1. Where can I get the NCERT Solutions Class 12 Mathematics Chapter 2 online?

2. How many questions are included in NCERT Solutions for Class 12 Mathematics Chapter 2?

3. What are inverse trigonometric functions?

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NCERT Solutions Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

Key Topics Covered In NCERT Solution for Class 12 Mathematics Chapter 2

Download NCERT Solutions for Class 12 Mathematics Chapter 2 Exercise & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of NCERT Solutions Class 12 Mathematics Chapter 2

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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